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      四种最简真分式的积分
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        <p>有理函数的原函数一定是初等函数，因此，理论上说有理函数的积分是一定能算出来的。有理函数可以写成多项式的商的形式，且根据多项式除法知，假分式=多项式+真分式，而多项式的积分是容易计算的，根据代数学有关知识，真分式又可化为四种最简真分式之和。因此，有理函数的积分可以归结为四种最简真分式的积分。<br><a id="more"></a><br>最简真分式只有以下四种：</p>
<script type="math/tex; mode=display">
\begin{align}
(1)&\frac{A}{x-a}\\
(2)&\frac{A}{(x-a)^m}\qquad\quad (m>1)\\
(3)&\frac{Mx+N}{x^2+px+q}\qquad (p^2-4q<0)\\
(4)&\frac{Mx+N}{(x^2+px+q)^k}\quad (k>1,p^2-4q<0)
\end{align}</script><p>下面讨论这四种积分：</p>
<h2 id="1"><a href="index.html#1" class="headerlink" title="1"></a>1</h2><script type="math/tex; mode=display">
\int \frac{A}{x-a}dx=A\ln|x-a|+C</script><h2 id="2"><a href="index.html#2" class="headerlink" title="2"></a>2</h2><script type="math/tex; mode=display">
\int \frac{A}{(x-a)^m}
=A\int(x-a)^{-m}d(x-a)
=\frac{A}{1-m}(x-a)^{1-m}+C</script><h2 id="3"><a href="index.html#3" class="headerlink" title="3"></a>3</h2><script type="math/tex; mode=display">
\begin{align}
\int \frac{Mx+N}{x^2+px+q}dx
&=\frac{M}{2}\int \frac{d(x^2+px+q)}{x^2+px+q}+\int \frac{(N-\frac{Mp}{2})dx}{x^2+px+q}\\
&=\frac{M}{2}\ln(x^2+px+q)+(N-\frac{Mp}{2})\int\frac{dx}{(x+\frac{p}{2})^2+\frac{4q-p^2}{4}}\\
&=\frac{M}{2}\ln(x^2+px+q)+\frac{N-\frac{Mp}{2}}{\frac{4q-p^2}{4}}\cdot\sqrt{\frac{4q-p^2}{4}}\int\frac{d(\frac{x+\frac{p}{2}}{\sqrt{\frac{4q-p^2}{4}}})}{(\frac{x+\frac{p}{2}}{\sqrt{\frac{4q-p^2}{4}}})^2+1}\\
&=\frac{M}{2}\ln(x^2+px+q)+\frac{2N-Mp}{\sqrt{4q-p^2}}arctan\frac{2x+p}{\sqrt{4q-p^2}}+C\\
\end{align}</script><h2 id="4"><a href="index.html#4" class="headerlink" title="4"></a>4</h2><script type="math/tex; mode=display">
\begin{align}
\int\frac{Mx+N}{(x^2+px+q)^k}
&=\frac{M}{2}\int\frac{d(x^2+px+q)}{(x^2+px+q)^k}+\int\frac{(N-\frac{Mp}{2})dx}{\left[(x+\frac{p}{2})^2+\frac{4q-p^2}{4}\right]^k}\\
&=\frac{M}{2(1-k)}(x^2+px+q)^{1-k}+(N-\frac{Mp}{2})\int\frac{dx}{\left[(x+\frac{p}{2})^2+\frac{4q-p^2}{4}\right]^k}
\end{align}</script><p>不妨设$t=x+\frac{p}{2},a=\sqrt{\frac{4q-p^2}{4}}$<br>则有</p>
<script type="math/tex; mode=display">
\int\frac{dx}{\left[(x+\frac{p}{2})^2+\frac{4q-p^2}{4}\right]^k}
=\int\frac{dt}{(t^2+a^2)^k}</script><p>设$I_k=\int\frac{dt}{(t^2+a^2)^k}$<br>则有</p>
<script type="math/tex; mode=display">
\begin{align}
I_k&=\int\frac{dt}{(t^2+a^2)^k}\\
&=\frac{t}{(t^2+a^2)^k}-\int td\left[\frac{1}{(t^2+a^2)^k}\right]\\
&=\frac{t}{(t^2+a^2)^k}+2k\int \frac{t^2+a^2-a^2}{(t^2+a^2)^{k+1}}dt\\
&=\frac{t}{(t^2+a^2)^k}+2k\int \frac{dt}{(t^2+a^2)^{k}}-2ka^2\int\frac{dt}{(t^2+a^2)^{k+1}}
\end{align}</script><p>即有递推式</p>
<script type="math/tex; mode=display">
I_k=\frac{t}{(t^2+a^2)^k}+2kI_k-2ka^2I_{k+1}\\</script><p>因此对于第四种情况，可以通过以下递推式计算。</p>
<script type="math/tex; mode=display">
\left\{
\begin{array}{}
I_{k+1}=\frac{2k-1}{2ka^2}I_k+\frac{t}{2ka^2(t^2+a^2)^k}\\
I_1=\frac{1}{a}arctan\frac{t}{a}+C
\end{array}
\right.</script><p>如有错误，望在评论区指出。<br>参考资料：<br>《工科数学分析第三版上册》华中科技大学出版社<br><a href="https://www.zybang.com/question/360235fb5c1b8715f822669e02490bed.html" target="_blank" rel="noopener">https://www.zybang.com/question/360235fb5c1b8715f822669e02490bed.html</a><br><a href="https://wenku.baidu.com/view/45280e1010a6f524ccbf85cf.html" target="_blank" rel="noopener">https://wenku.baidu.com/view/45280e1010a6f524ccbf85cf.html</a></p>

      
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